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y ′′′ − 6y ′′ + 11y ′ − 6y = 0 λ 3 − 6λ 2 +
11λ − 6 = 0 −6
±1, ±2, ±3, ±6 λ1 = 1 λ2 = 2 λ3 = 3
x 2
0 ¯y(x)
2 ¯y(x) = ax 2 + bx + c
−12a + 11(2ax + b) − 6(ax 2 + bx + c) = x 2
a = −1/6 2 − 11x/3 + 11b − 6bx − 6c = 0 6b = −11/3 b = −11/18
6c = −85/18 c = −85/108
y(x) = c1e x + c2e 2x + c3e 3x − x2
6
11 85
− x −
18 108 ,
c1, c2, c3 ∈ R
y IV − y ′′ = 0 λ 4 − λ 2 = 0
0 2 ±1 x − 1
0 2
¯y(x) = x 2 (ax + b) = ax 3 + bx 2
−(6ax + 2b) = x − 1 a = −1/6 b = 1/2
y(x) = c1 + c2x + c3e x + c4e −x + x 2
1 x
− .
2 6
k ∈ N
− 1
k u′′ (x) + |u ′ (x)| = 1, x ∈ (0, 1),
u(0) = 0, u(1) = 2.
uk
{uk} [0, 1]
v(x) = u ′ (x) v ′ (x) = k(|v(x)| − 1)
v(x) ≡ 1 v(x) ≡ −1
v(x) < −1 v(x) = 1 −1 < v(x) < 1
v(x) = 1 v(x) > 1 x ∈ (0, 1)
u(x) = u(0) +
x
0
v(t) dt,
u(0) = 0 u(1) = 2
2 =
1
0
v(t) dt,
v(x) = u ′ (x) > 1
v ′ (x) = kv(x) − k
u(x)
u(x) = u(0) +
x
0
v(t) = ce kt + 1.
(ce kt + 1) dx = u(0) + c
k [ekt ] t=x
t=0 + x = u(0) + c
k (ekx − 1) + x.